Sine and Cosine help please!

[ChickySeeChickyDo]

I'm working on an assignment with 22 questions and I'm completely baffled by 2 of them. D: I don't want the answers, I just need to be told how to get the answer.

1) Imagine a right triangle ACB /_| (my attempt at making a triangle XD) A is on the far left, C is on the far right and the right angle, and B is on top. Ang A is 45 degrees, and AB = 30. What is the length of AC?

I know that one is Cos45 but I can't figure out the answer. I thought you put Cos45/1 = 30/x and you divide Cos45 by 30 but what I came up with wasn't any of the answer choices.

2) Imagine a right triangle ACB /_| A is on the far left, C is on the far right and the right angle, and B is on top. Ang A is 45 degrees, and AB = 30. What is the length of BC?

I know this one is Sin45 and I tried the same method as I did the other one but again, what I came up with wasn't an answer choice. :/

 

[BigGlenntheHeavy]

\(\displaystyle 1) \ cos(45^0) \ = \ \frac{AC}{30}\)

\(\displaystyle 2) \ sin(45^0) \ = \ \frac{BC}{30}\)

\(\displaystyle Now, \ cross-multiply \ and \ you \ got \ it.\)

\(\displaystyle Also, \ note, \ you \ have \ an \ isosceles \ right \ triangle, \ (AC \ = \ BC)\)

\(\displaystyle Hence, \ let \ x \ = \ AC \ or \ BC, \ then \ x^2 \ = \ \frac{30^2}{2}, \ solve \ for \ x\)

 

[soroban]

Hello, ChickySeeChickyDo!

We can solve this without Trigonometry . . .

1) Given: right triangle \(\displaystyle ABC\!:\;\;\angle A = 45^o,\;\angle C = 90^o,\; AB = 30.\)
What is the length of \(\displaystyle AC\)?

                  * B
                * |
              *45d|
        30  *     | x
          *       |
        * 45d     |
    A * - - - - - * C
            x

Since \(\displaystyle \angle A \,=\,\angle B \,=\,45^o\), the triangle is isosceles.
Hence: .\(\displaystyle AC \,=\,BC\,=\,x\)

Pythagorus says: .\(\displaystyle x^2+x^2 \:=\:30^2 \quad\Rightarrow\quad 2x^2 \:=\:900 \quad\Rightarrow\quad x^2 \:=\:450\)

Therefore: .\(\displaystyle x \:=\:\sqrt{450} \:=\:15\sqrt{2}\)